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Find the Value of P for Which the Quadratic Equation ( P + 1 ) X 2 − 6 ( P + 1 ) X + 3 ( P + 9 ) = 0 , P ≠ − 1 Has Equal Roots. Hence, Find the Roots of the Equation. - CBSE Class 10 - Mathematics

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Question

Find the value of p for which the quadratic equation 

\[\left( p + 1 \right) x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq - 1\] has equal roots. Hence, find the roots of the equation.

Disclaimer: There is a misprinting in the given question. In the question 'q' is printed instead of 9.

Solution

The given quadratic equation  \[\left( p + 1 \right) x^2 - 6(p + 1)x + 3(p + 9) = 0\],

has equal roots.

Here, 

\[a = p + 1, b = - 6p - 6 \text { and } c = 3p + 27\].

As we know that \[D = b^2 - 4ac\]

Putting the values of \[a = p + 1, b = - 6p - 6\text { and } c = 3p + 27\].

\[D = \left[ - 6(p + 1) \right]^2 - 4\left( p + 1 \right)\left[ 3\left( p + 9 \right) \right]\]

\[ = 36( p^2 + 2p + 1) - 12( p^2 + 10p + 9)\]

\[ = 36 p^2 - 12 p^2 + 72p - 120p + 36 - 108\]

\[ = 24 p^2 - 48p - 72\]

The given equation will have real and equal roots, if D = 0

Thus, 

\[24 p^2 - 48p - 72 = 0\]

\[\Rightarrow p^2 - 2p - 3 = 0\]

\[ \Rightarrow p^2 - 3p + p - 3 = 0\]

\[ \Rightarrow p(p - 3) + 1(p - 3) = 0\]

\[ \Rightarrow (p + 1)(p - 3) = 0\]

\[ \Rightarrow p + 1 = 0 \text { or } p - 3 = 0\]

\[ \Rightarrow p = - 1 \text { or } p = 3\]

It is given that p ≠ −1, thus p = 3 only.
Now the equation becomes

\[4 x^2 - 24x + 36 = 0\]

\[ \Rightarrow x^2 - 6x + 9 = 0\]

\[ \Rightarrow x^2 - 3x - 3x + 9 = 0\]

\[ \Rightarrow x(x - 3) - 3(x - 3) = 0\]

\[ \Rightarrow (x - 3 )^2 = 0\]

\[ \Rightarrow x = 3, 3\]

​Hence, the root of the equation is 3.

  Is there an error in this question or solution?

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Solution Find the Value of P for Which the Quadratic Equation ( P + 1 ) X 2 − 6 ( P + 1 ) X + 3 ( P + 9 ) = 0 , P ≠ − 1 Has Equal Roots. Hence, Find the Roots of the Equation. Concept: Solutions of Quadratic Equations by Factorization.
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